w=lnsecx+tgx dw= 1/secx+tgx * (-1/cosx^2 *-senx +secx^2)dx= dw= cosxdx cosxe^w-1=rq(1-cosx^2) e^2wcosx^2-2cosxe^w+1=1-cosx^2 cosx^2(e^2w+1)-2cosxe^w=0 cosx= 2e^w/(e^2w+1)=2/(e^w+e^-w)=1/coshw I w *coshw dw u= w du=dw dv=coshwdw v= senhw I ln(sec x+tgx)dx= w*senhw-Isenhwdw= wsenhw-coshw =`(n(secx+tgx) -1/cosx
On 7/31/07, antonio ricardo <[EMAIL PROTECTED]> wrote: > > ola > poderiam me ajudar a resolver a seguinte integral > > integral de ln(secx + tgx) > > valeu > > Alertas do Yahoo! Mail em seu celular. Saiba > mais<http://br.mobile.yahoo.com/mailalertas/>. > >
